Answer
See the explanation
Work Step by Step
Let's determine the electron configurations for the most stable ions formed by the given elements:
1. Aluminum (Al):
Atomic number of Al is 13, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^1.
The most stable ion formed by Al is Al^3+, which has the electron configuration 1s^2 2s^2 2p^6.
2. Barium (Ba):
Atomic number of Ba is 56, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 5s^2.
The most stable ion formed by Ba is Ba^2+, which has the electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6.
3. Selenium (Se):
Atomic number of Se is 34, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^4.
The most stable ion formed by Se is Se^2-, which has the electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6.
4. Iodine (I):
Atomic number of I is 53, and its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^5.
The most stable ion formed by I is I^-, which has the electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6.
